Introduction

In this blog post, we will explore how to solve the LeetCode problem "1497" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an array of integers arr of even length n and an integer k. We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k. Return true If you can find a way to do that or false otherwise. Example 1: Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5 Output: true Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10). Example 2: Input: arr = [1,2,3,4,5,6], k = 7 Output: true Explanation: Pairs are (1,6),(2,5) and(3,4). Example 3: Input: arr = [1,2,3,4,5,6], k = 10 Output: false Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10. Constraints: arr.length == n 1 <= n <= 105 n is even. -109 <= arr[i] <= 109 1 <= k <= 105

Explanation

Here's the breakdown of the solution:

• Residue Counting: The core idea is to count the frequency of each residue (remainder) when the numbers in the array are divided by k.
• Pair Matching: For each residue r, we need to find a corresponding residue k - r such that their counts match. If the count of r does not equal the count of k-r, it's impossible to form the required pairs. Special handling is needed for the residue 0 and k/2 if k is even because their complement is themselves.

• Early Exit: If, at any point, we find a mismatch in the counts, we can immediately return False.

• Runtime Complexity: O(n), Storage Complexity: O(k)

Code

    def can_arrange(arr: list[int], k: int) -> bool:
    """
    Given an array of integers arr of even length n and an integer k.
    We want to divide the array into exactly n / 2 pairs such that the sum of each pair is divisible by k.
    Return true If you can find a way to do that or false otherwise.

    Example 1:
    Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5
    Output: true
    Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).

    Example 2:
    Input: arr = [1,2,3,4,5,6], k = 7
    Output: true
    Explanation: Pairs are (1,6),(2,5) and(3,4).

    Example 3:
    Input: arr = [1,2,3,4,5,6], k = 10
    Output: false
    Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.

    Constraints:
    arr.length == n
    1 <= n <= 105
    n is even.
    -109 <= arr[i] <= 109
    1 <= k <= 105
    """
    remainders = {}
    for num in arr:
        remainder = num % k
        remainders[remainder] = remainders.get(remainder, 0) + 1

    for remainder in remainders:
        if remainder == 0:
            if remainders[remainder] % 2 != 0:
                return False
        elif 2 * remainder == k:
            if remainders[remainder] % 2 != 0:
                return False
        elif remainder not in remainders or (k - remainder) not in remainders or remainders[remainder] != remainders[k - remainder]:
            return False
    return True